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An ideal gas undergoes a reversible isochoric process. The initial pressure is $2 \mathrm{~atm}$ , and the initial temperature is $300 \mathrm{~K}$ . During the process, $800 \mathrm{~J}$ of heat is added to the gas. Calculate the final temperature of the gas. Given the heat capacity at constant volume $\mathrm{\left(C_v\right) \: is\: 20 \mathrm{~J} / \mathrm{K}.}$

Option: 1
$200 \mathrm{~K}$

Option: 2
$300 \mathrm{~K}$

Option: 3
$240 \mathrm{~K}$

Option: 4
$340 \mathrm{~K}$

Answers (1)

best_answer

Given data:

Initial pressure $\mathrm{\left(P_i\right)=2 atm}$

Initial temperature $\mathrm{\left(T_i\right)=300 \mathrm{~K}}$

Heat added $\mathrm{(Q)=800 \quad \mathrm{~J}}$

Heat capacity at constant volume $\mathrm{\left(C_v\right)=20 \mathrm{~J} / \mathrm{K}}$

The change in internal energy $\mathrm{(\Delta U)}$ during an isochoric process is given by:

$\mathrm{ \Delta U=Q=n C_v \Delta T }$

Solving for the change in temperature $\mathrm{ (\Delta T): }$

$\mathrm{ \Delta T=\frac{Q}{n C_v} }$

Substitute the values and calculate $\mathrm{ \Delta T : }$

$\mathrm{ \Delta T=\frac{800 \mathrm{~J}}{n \cdot 20 \mathrm{~J} / \mathrm{K}}=\frac{800 \mathrm{~J}}{1 \mathrm{~mol} \cdot 20 \mathrm{~J} / \mathrm{K}}=40 \mathrm{~K} }$

The final temperature $\mathrm{ \left(T_f\right) }$ can be calculated by adding $\mathrm{ \Delta T }$ to the initial temperature $\mathrm{ \left(T_i\right) : }$
$\mathrm{ T_f=T_i+\Delta T=300 \mathrm{~K}+40 \mathrm{~K}=340 \mathrm{~K} }$

Therefore, the final temperature of the gas is $\mathrm{ 340 \mathrm{~K}. }$

So correct option is 4.

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